Differentiaali- ja integraalilaskenta

Huom. Koska \(\mathbb{N}\) on järjestetty joukko, niin myös jonon termeillä \( f(n)\) on vastaava järjestys. Sen sijaan joukon alkioilla ei yleisessä tapauksessa ole määrättyä järjestystä.

 Funktion \[\begin{aligned} f:\mathbb{N} \rightarrow & M \\ n \mapsto & a_{n}\end{aligned}\] perusteella jokaiseen jonon termiin liittyy yksikäsitteinen lukua \(n\in\mathbb{N}\) to each term. Se merkitään alaindeksinä ja sitä kutsutaan vastaavan jonon termin indeksiksi; jokainen jonon termi voidaan siis tunnistaa sen indeksin avulla.

Suppeneminen

Jonosta \((a_k)\) voidaan muodostaa sen osasummia asettamalla \[s_n =a_1+a_2+\dots+a_n.\]

Jos osasumminen jonolla \((s_n)\) on raja-arvo \(s\in \mathbb{R}\), niin luvuista \((a_k)\) muodostettu sarja suppenee ja sen summa on \(s\). Tällöin merkitään \[ a_1+a_2+\dots =\sum_{k=1}^{\infty} a_k = \lim_{n\to\infty}\underbrace{\sum_{k=1}^{n} a_k}_{=s_{n}} = s. \]

Indeksöinti

Osasummat kannattaa indeksöidä samalla tavalla kuin alkuperäinen jono \((a_k)\); esimerkiksi jonon \((a_k)_{k=0}^{\infty}\) osasummat ovat \(s_0= a_0, s_1=a_0+a_1\) jne.

Sarjaan voidaan tehdä indeksinsiirtoja ilman että varsinainen sarja muuttuu: \[\sum_{k=1}^{\infty} a_k =\sum_{k=0}^{\infty} a_{k+1} = \sum_{k=2}^{\infty} a_{k-1}.\]

Konkreettinen esimerkki: \[\sum_{k=1}^{\infty} \frac{1}{k^2}=1+\frac{1}{4}+\frac{1}{9}+\dots= \sum_{k=0}^{\infty} \frac{1}{(k+1)^2}\]

Sarjan hajaantuminen

Sarja, joka ei suppene, on hajaantuva. Tämä voi tapahtua kolmella eri tavalla:

1. sarjan osasummat kasvavat rajatta kohti ääretöntä
   
2. sarjan osasummat pienenevät rajatta kohti miinus ääretöntä
   
3. osasummien jono heilahtelee niin, ettei sillä ole raja-arvoa.

Hajaantuvan sarjan kohdalla merkintä \(\displaystyle\sum_{k=1}^{\infty} a_k\) ei oikeastaan tarkoita mitään (se ei ole reaaliluku). Tällöin voidaan tulkita, että merkintä tarkoittaa osasummien jonoa, joka on aina hyvin määritelty.

Geometrinen sarja

Geometrinen sarja \[\sum_{k=0}^{\infty} aq^k\] suppenee, jos \(|q|<1\) (tai \(a=0\)), jolloin sen summa on \(\frac{a}{1-q}\). Jos \(|q|\ge 1\), niin sarja hajaantuu.

Yleisemmin \[\sum_{k=i}^{\infty} aq^k = \frac{aq^i}{1-q} = \frac{\text{sarjan 1. termi}}{1-q},\text{ jos } |q|<1.\]

Esimerkki 1.

Määritä sarjan \[\sum_{k=1}^{\infty}\frac{3}{4^{k+1}}\] summa.

Ratkaisu. Koska \[\frac{3}{4^{k+1}} = \frac{3}{4}\cdot \left( \frac{1}{4}\right)^k,\] niin kyseessä on geometrinen sarja. Sen summa on \[\frac{3}{4}\cdot \frac{1/4}{1-1/4} = \frac{1}{4}.\]

Laskusääntöjä

• \(\displaystyle{\sum_{k=1}^{\infty} (a_k+b_k) = \sum_{k=1}^{\infty} a_k + \sum_{k=1}^{\infty} b_k}\)
• \(\displaystyle{\sum_{k=1}^{\infty} (c\, a_k) = c\sum_{k=1}^{\infty} a_k}\), kun \(c\in \mathbb{R}\) on vakio
  

Huomautus: Sarjoilla ei ole jonojen kaltaista tulosääntöä, koska jo kahden termin summille \[(a_1+a_2)(b_1+b_2) \neq a_1b_1 +a_2b_2.\] Tulosäännön oikea muoto on sarjojen Cauchy-tulo, jossa myös ristitermit otetaan huomioon.

Katso esimerkiksi https://en.wikipedia.org/wiki/Cauchy_product

Todistus.

Huomautus: Ominaisuutta \(\lim_{k\to \infty} a_k = 0\) ei voi käyttää sarjan suppenemisen osoittamiseen; vrt. seuraavat esimerkit. Tämä on eräs yleisimmistä päättelyvirheistä sarjojen kohdalla!

Esimerkki

Tutki sarjan \[\sum_{k=1}^{\infty} \frac{k}{k+1} = \frac{1}{2}+\frac{2}{3}+\frac{3}{4}+\dots\] suppenemista.

Ratkaisu. Sarjan yleisen termin raja-arvo on \[\lim_{k\to\infty}\frac{k}{k+1} = 1.\] Koska raja-arvo ei ole nolla, niin sarja hajaantuu.

Harmoninen sarja

Harmoninen sarja \[\sum_{k=1}^{\infty} \frac{1}{k} = 1+\frac{1}{2}+\frac{1}{3}+\dots\] hajaantuu, vaikka yleisen termin \(a_k=1/k\) raja-arvo on nolla.

Todistus.

Positiiviset sarjat

Sarjan summan laskeminen on usein vaikeata ja monesti jopa mahdotonta, jos vaatimuksena on summan eksplisiittinen lauseke. Moniin sovelluksiin riittää myös summan likiarvo, mutta sitä ennen olisi syytä selvittää, onko sarja suppeneva vai hajaantuva.

Sarja \(\displaystyle{\sum_{k=1}^{\infty} p_k}\) on positiivinen (tai positiiviterminen), jos \(p_k > 0\) kaikilla \(k\).

Positiivisten sarjojen suppeneminen on hyvin selväpiirteistä:

Esimerkki

Osoita, että superharmonisen sarjan \[\sum_{k=1}^{\infty}\frac{1}{k^2}\] osasummille pätee \(s_n<2\) kaikilla \(n\), joten sarja suppenee.

Ratkaisu. Ratkaisu perustuu epäyhtälöön \[\frac{1}{k^2} < \frac{1}{k(k-1)} = \frac{1}{k-1}-\frac{1}{k},\] kun \(k\ge 2\), koska sen mukaan \[\sum_{k=1}^n\frac{1}{k^2} < 1+ \sum_{k=2}^n\frac{1}{k(k-1)} =2-\frac{1}{n}< 2\] kaikilla \(n\ge 2\).

Tämän päättelyn voi tehdä myös integraalin avulla.

Leonhard Euler selvitti vuonna 1735, että sarjan summa on \(\pi^2/6\). Perusteluna hän käytti sinifunktion sarja- ja tulokehitelmien vertailua.

Todistus.

Esimerkki

Tutki vuorottelevan (= etumerkit vaihtelevat vuorotellen + ja -) sarjan \[\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k^2}=1-\frac{1}{4}+\frac{1}{9}-\dots\] suppenemista.

Ratkaisu. Koska \[\displaystyle{\left| \frac{(-1)^{k+1}}{k^2}\right| =\frac{1}{k^2}}\] ja superharmoninen sarja \[\sum_{k=1}^{\infty}\frac{1}{k^2}\] suppenee, niin tutkittava sarja suppenee itseisesti, ja sen vuoksi myös tavallisessa mielessä.

Vuorotteleva harmoninen sarja

Itseinen suppeneminen ei kuitenkaan tarkoita samaa kuin tavallinen suppeneminen:

Esimerkki

Vuorotteleva harmoninen sarja \[\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k} = 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\dots\] suppenee, mutta ei itseisesti.

(Idea) Piirretään osasummajonon \((s_n)\) kuvaaja, josta huomataan, että parillisten ja parittomien indeksien osasummat \(s_{2n}\) ja \(s_{2n+1}\) ovat monotonisia ja suppenevat kohti samaa raja-arvoa.

Sarjan summa on \(\ln 2\), joka saadaan selville integroimalla geometrisen sarjan summakaavaa.

Vertailutestit

Edelliset tarkastelut yleistyvät seuraavalla tavalla:

Todistus.

Esimerkki

Tutki sarjojen \[ \sum_{k=1}^{\infty} \frac{1}{1+k^3} \ \text{ ja }\ \sum_{k=1}^{\infty} \frac{1}{\sqrt{k}} \] suppenemista.

Ratkaisu. Koska\[0<\frac{1}{1+k^3} < \frac{1}{k^3}\le \frac{1}{k^2}\] kaikilla \(k\in \mathbb{N}\), niin ensimmäinen sarja suppenee majoranttiperiaatteen nojalla.

Toisaalta \[\displaystyle{\frac{1}{\sqrt{k}}\ge \frac{1}{k}}\] kaikilla \(k\in\mathbb{N}\), joten toisella sarjalla on hajaantuva harmoninen minorantti, joten se hajaantuu.

Suhdetesti

Käytännössä eräs parhaista tavoista tutkia suppenemista on suhdetesti, jossa sarjan peräkkäisten termien käyttäytymistä verrataan sopivaan geometriseen sarjaan:

Todistus.

Suhdetestin raja-arvomuoto

(Idea) Geometriselle sarjalle kahden peräkkäisen termin suhde on täsmälleen \(q\). Suhdetestin mukaan muidenkin sarjojen suppenemista voidaan (usein) tutkia samalla periaatteella, kun suhdeluku \(q\) korvataan tällä raja-arvolla.

Todistus.

For a subset of real numbers, denoted by \(S\), assume that \(x_0\) is such point that there is a sequence of points \((x_k)\in S\) such that \(x_k\to x_0\) as \(k\to \infty\). Here the set \(S\) is often the set of all real numbers, but sometimes an interval (open or closed).

Example 1.

Note that it is not necessary for \(x_0\) to be in \(S\). For example, the sequence \(x_k = 1/k\to 0\) as \(k\to \infty\) in \(S=]0,2[\), and \(x_k\in S\) for all \(k=1,2,\ldots\) but \(0\) is not in \(S\).

We consider a function \(f\) defined in the set \(S\). Then we define the limit of the function \(f\colon S\to \mathbb{R}\) at \(x_0\) as follows.

The function \(f\colon \mathbb{R} \to \mathbb{R}\) defined by \(f(x)=x^2\) has a limit \(0\) at the point \(x=0\).

The function \(g\colon\mathbb{R}\to \mathbb{R}\) defined by \[g(x)= \left\{\begin{array}{rl}0 & \text{ for }x<0, \\ 1 & \text{ for }x\ge 0.\end{array}\right.\] does not have a limit at the point \(x=0\). To formally prove this, take sequences \((x_k)\), \((y_k)\) defined by \(x_k=1/k\) and \(y_k=-1/k\) for \(k=1,2,\ldots\). Then the both sequences are in \(S=\mathbb{R}\), but \(f(x_k)=1\) and \(f(y_k)=0\) for any \(k\).

Example 0.

The graph below shows how far a cyclist gets from his starting point.

a) Look at the red line. We can see that in three hours, the cyclist moved \(20\)km. The average speed of the whole trip is \(6.6\) km/h.
b) Now look at the green line. We can see that during the third hour the cyclist moved \(10\)km further. That makes the average speed of that time interval \(10\) km/h.
Notice that the slope of the red line is \(20/3 \approx 6.6\) and that the slope of the blue line is \(10\). These are the same values as the corresponding average speeds.
c) Look at the blue line. It is the tangent of the curve at the point \(x=2h\). Using the same principle as with average speeds, we conclude that after two hours of the departure, the speed of the cyclist was \(30/2\) km/h \(= 15\) km/h.

Now we will proceed to the general definition:

Note: Since \(x = x_0+h\), then \(h=x-x_0\), and thus the definition can also be written in the form \[f'(x_0):=\lim_{x\to x_0} \frac{f(x)-f(x_0)}{x-x_0}.\]

The derivative can be denoted in different ways: \[ f'(x_0)=Df(x_0) =\left. \frac{df}{dx}\right|_{x=x_0}, \ \ f'=Df =\frac{df}{dx}. \]

Interpretation. Consider the curve \(y = f(x)\). Now if we draw a line through the points \((x_0,f(x_0))\) and \((x_0+h, f(x_0+h))\), we see that the slope of this line is \[\frac{f(x_0+h)-f(x_0)}{x_0+h-x_0} = \frac{f(x_0+h)-f(x_0)}{h}.\] When \(h \to 0\), the line intersects with the curve \(y = f(x)\) only in the point \((x_0, f(x_0))\). This line is the tangent of the curve \(y=f(x)\) at the point \((x_0,f(x_0))\) and its slope is \[\lim_{h\to 0} \frac{f(x_0+h)-f(x_0)}{h},\] which is the derivative of the function \(f\) at \( x_0\). Hence, the tangent is given by the equation \[y=f(x_0)+f'(x_0)(x-x_0).\]

Interactivity. Move the point of intersection and observe changes on the tangent line of the curve.

Let \(f\colon \mathbb{R} \to \mathbb{R}\) be the function \(f(x) = x^3 + 1\). The derivative of \(f\) at \(x_0 = 1\) is \[\begin{aligned}f'(1) &=\lim_{h \to 0} \frac{f(1+h)-f(1)}{h} \\ &=\lim_{h \to 0} \frac{(1+h)^3 + 1 - 1^3 - 1}{h} \\ &=\lim_{h \to 0} \frac{1+3h+3h^2+h^3-1}{h} \\ &=\lim_{h \to 0} \frac{h(3+3h+h^2)}{h} \\ &=\lim_{h \to 0} 3+3h+h^2 \\ &= 3. \end{aligned}\]

Let \(f\colon \mathbb{R} \to \mathbb{R}\) be the function \(f(x)=ax+b\). We find the derivative of \(f(x)\).

Immediately from the definition we get: \[\begin{aligned}f'(x) &=\lim_{h\to 0} \frac{f(x+h)-f(x)}{h} \\ &=\lim_{h\to 0} \frac{[a(x+h)+b]-[ax+b]}{h} \\ &=\lim_{h\to 0} a \\ &=a.\end{aligned}\]

Here \(a\) is the slope of the tangent line. Note that the derivative at \(x\) does not depend on \(x\) because \(y=ax+b\) is the equation of a line.

Note. When \(a=0\), we get \(f(x) = b\) and \(f'(x) = 0\). The derivative of a constant function is zero.

Let \(g\colon \mathbb{R} \to \mathbb{R}\) be the function \(g(x)=|x|\). Does \(g\) have a derivative at \(0\)?

Now \[g'(x_0)= \begin{cases}+1 & \text{when $x_{0}>0$} \\ -1 & \text{when $x_{0}<0$}\end{cases}\]

The graph \(y=g(x)\) has no tangent at the point \(x_0=0\): \[\frac{g(0+h)-g(0)}{h}= \frac{|0+h|-|0|}{h}=\frac{|h|}{h}=\begin{cases}+1 & \text{for $h>0$}, \\ -1 & \text{for $h<0$}.\end{cases}\] Thus \(g'(0)\) does not exist.

Conclusion. The function \(g\) is not differentiable at the point \(0\).

Remark. Let \(f\colon (a,b)\to \mathbb{R}\). If \(f'(x)\) exists for every \(x\in (a,b)\) then we get a function \(f'\colon (a,b)\to \mathbb{R}\). We write:

Here \(f''(x)\) is called the second derivative of \(f\) at \(x\), \(f^{(3)}\) is the third derivative, and so on.

We introduce the notation \begin{eqnarray} C^n\bigl( ]a,b[\bigr) =\{ f\colon \, ]a,b[\, \to \mathbb{R} & \mid & f \text{ is } n \text{ times differentiable on the interval } ]a,b[ \nonumber \\ & & \text{ and } f^{(n)} \text{ is continuous}\}. \nonumber \end{eqnarray} These functions are said to be n times continuously differentiable.

The distance moved by a cyclist (or a car) is given by \(s(t)\). Then the speed at the moment \(t\) is \(s'(t)\) and the acceleration is \(s''(t)\).

If \(f\) is differentiable at the point \(x_0\), then \(f\) is continuous at the point \(x_0\): \[ \lim_{h\to 0} f(x_0+h) = f(x_0).\] Why? Because if \(f\) is differentiable, then we get \[f(x_0)+h\frac{f(x_0+h)-f(x_0)}{h} \rightarrow f(x_0)+0\cdot f'(x_0)=f(x_0),\] as \(h \to 0\).

Note. If a function is continuous at the point \(x_0\), it doesn't have to be differentiable at that point. For example, the function \(g(x) = |x|\) is continuous, but not differentiable at the point \(0\).

Next we will give some important rules which are often applied in practical problems concerning determination of the derivative of a given function.

Suppose that \(f\) and \(g\) are differentiable at \(x\).

Proof.

Proof.

Proof.

Proof.

Proof.

Proof.

Example 1.

\[\frac{d}{dx}(x^{2006}+5x^3+42)=\frac{d}{dx}x^{2006}+5\frac{d}{dx}x^3+42\frac{d}{dx}1=2006x^{2005}+5\cdot 3x^2.\]

\[\begin{aligned}\frac{d}{dx} [(x^4-2)(2x+1)] &= \frac{d}{dx}(x^4-2) \cdot (2x+1) + (x^4-2) \cdot \frac{d}{dx}(2x + 1) \\ &= 4x^3(2x+1) + 2(x^4-2) \\ &= 8x^4+4x^3+2x^4-4 \\ &= 10x^4+4x^3-4.\end{aligned}\]

Note. We can check the answer by deriving it in another way: \[\frac{d}{dx} [(x^4-2)(2x+1)] = \frac{d}{dx} (2x^5 +x^4 -4x -2) = 10x^4 +4x^3 -4.\]

For \(x \neq 0\) we get \[\frac{d}{dx} \frac{3}{x^3} = 3 \cdot \frac{d}{dx} \frac{1}{x^3} = -3 \cdot \frac{\frac{d}{dx} x^3}{(x^3)^2} = -3 \cdot \frac{3x^2}{x^6}= - \frac{9}{x^4}.\]

Note. There is another way of solving the problem above by noticing that \(\frac{1}{x^3} = x^{-3}\) and differentiating it as a power: \[\frac{d}{dx} \ \frac{3}{x^3} = 3 \cdot \frac{d}{dx} x^{-3} = 3 \cdot (-3x^{-4})= - \frac{9}{x^4}\]

\[\begin{aligned}\frac{d}{dx} \frac{x^3}{1+x^2} & = \frac{(\frac{d}{dx}x^3)(1+x^2)-x^3\frac{d}{dx}(1+x^2)}{(1+x^2)^2} \\ & = \frac{3x^2(1+x^2)-x^3(2x)}{(1+x^2)^2} \\ & = \frac{3x^2+x^4}{(1+x^2)^2}.\end{aligned}\]

Proof (idea).

Example 1.

\[\frac{d}{dx} (3 \sin(x)) = 3 \sin'(x) = 3 \cos(x).\]

Example 2.

\[\frac{d}{dx} \cos^2 (x) = \cos'(x) \cdot \cos(x) + \cos(x) \cdot \cos'(x) = -2\sin(x)\cos(x).\]

Example 3.

\[\begin{aligned} \frac{d}{dx} \frac{\sin(x) + 1}{\cos(x)} &= \frac{d}{dx} \left( \frac{\sin(x)}{\cos(x)} + \frac{1}{\cos(x)} \right) \\ &= \tan'(x) - \frac{\cos'(x)}{\cos^2(x)} \\ &= \frac{1+\sin(x)}{\cos^2 (x)}.\end{aligned}\]

The Chain Rule.

The problem is to differentiate the function \((2x-1)^3\). We take \(f(x) = x^3\) and \(g(x) = 2x-1\) and differentiate the composite function \(f(g(x))\). As \[f'(x) = 3x^2 \text{ and } g'(x) = 2,\] we get \[\frac{d}{dx} (2x-1)^3 = 3(2x-1)^2 \cdot 2 = 6(4x^2-4x+1) = 24x^2-24x+6.\]

We need to differentiate the function \(\sin 3x\). Take \(f(x) = \sin x\) and \(g(x) = 3x\), then differentiate the composite function \(f(g(x))\). \[\frac{d}{dx} \sin 3x = \cos 3x \cdot 3 = 3 \cos 3x.\]

Remark. Let \(h\colon \mathbb{R}\to \mathbb{R}, g\colon \mathbb{R}\to \mathbb{R}\) and \(f\colon \mathbb{R}\to \mathbb{R}\). Now \[\frac{d}{dx}f(g(h(x)))=f'(g(h(x)))\frac{d}{dx}g(h(x))=f'(g(h(x)))g'(h(x))h'(x).\] Similarly, one may obtain even more complex rules for composites of multiple functions.

Differentiate the function \(\cos^3 2x\). Take \(f(x) = x^3\), \(g(x) = \cos x\) and \(h(x) = 2x\) and differentiate the composite function \(f(g(h(x)))\). \[\begin{aligned}\frac{d}{dx} \cos^3 2x &= 3(\cos 2x)^2 \cdot \frac{d}{dx} \cos 2x \\ &= 3 \cos^2 2x \cdot (-\sin 2x) \cdot 2 \\ &= -6 \sin 2x \cos^2 2x.\end{aligned}\]

Remark. If \(x_0\) is a local maximum value and \(f'(x_0)\) exists, then \[\begin{cases}f'(x_0) & =\lim_{h\to 0^{+}}\frac{f(x_0+h)-f(x_0)}{h} \leq 0 \\ f'(x_0) & =\lim_{h\to 0^{-}}\frac{f(x_0+h)-f(x_0)}{h} \geq 0.\end{cases}\] Hence \(f'(x_0)=0\).

We get:

Let \(f: \mathbb{R} \to \mathbb{R}\) be defined by \[f(x) = x^3 -3x + 1.\] Then \[f'(x) = 3x^2-3\] and we can see that at the points \(x_0 = -1\) and \(x_0 = 1\) the local maximum and minimum of \(f\) are obtained, \[f'(-1) = 3 \cdot (-1)^2 - 3 = 0 \text{ and } f'(1) = 3 \cdot 1^2 - 3 = 0.\]

In practice, when we are looking for the local extrema of a given function, we need to check three kinds of points:

1. the zeros of the derivative

2. the endpoints of the domain of definition (interval)

3. points where the function is not differentiable

If we happened to know beforehand that the function has a minimum/maximum, then we start off by finding all the possible local extrema (the points described above), evaluate the function at these points and pick the greatest/smallest of these values.

Let us find the smallest and greatest value of the function \(f\colon [0,2]\to \mathbf{R}\), \(f(x)=x^3-6x\). Since the function is continuous on a closed interval, then it has a maximum and a minimum. Since the function is differentiable, it is sufficient to examine the endpoints of the interval and the zeros of the derivative that are contained in the interval.

The zeros of the derivative: \(f'(x)=3x^2-6=0 \Leftrightarrow x=\pm \sqrt{2}\). Since \(-\sqrt{2}\not\in [0,2]\), we only need to evaluate the function at three points, \(f(0)=0\), \(f(\sqrt{2})=-4\sqrt{2}\) and \(f(2)=-4\). From these we can see that the smallest value of the function is \(-4\sqrt{2}\) and the greatest value is \(0\), respectively.

Next we will formulate a fundamental result for differentiable functions. The basic idea here is that the change on an interval can only happen, if there is change at some point on the inverval.

Proof.

This result has an important application:

Proof.

For the polynomial \(f(x) = \frac{1}{4} x^4-2x^2-7\) the derivative is \[f'(x) = x^3-4x = x(x^2-4) = 0,\] when \(x=0\), \(x=2\) or \(x=-2\). Now we can draw a table:

We need to find a rectangle so that its area is \(9\) and it has the least possible perimeter.

Let \(x\ (>0)\) and \(y\ (>0)\) be the sides of the rectangle. Then \(x \cdot y = 9\) and we get \(y=\frac{9}{x}\). Now the perimeter is \[2x+2y = 2x+2 \frac{9}{x} = \frac{2x^2+18}{x}.\] In which point does the function \(f(x) = \frac{2x^2+18}{x}\) get its minimum value? Function \(f\) is continuous and differentiable, when \(x>0\) and using the quotient rule, we get \[f'(x) = \frac{4x \cdot x-(2x^2+18) \cdot 1}{x^2} = \frac{2x^2-18}{x^2}.\] Now \(f'(x) = 0\), when \[\begin{aligned}2x^2-18 &= 0 \\ 2x^2 &= 18 \\ x^2 &= 9 \\ x &= \pm 3\end{aligned}\] but we have defined that \(x>0\) and therefore are only interested in the case \(x=3\). Let's draw a table:

As the function \(f\) is continuous, we now know that it attains its minimum at the point \(x=3\). Now we calculate the other side of the rectangle: \(y=\frac{9}{x}=\frac{9}{3}=3\).

Thus, the rectangle, which has the least possible perimeter is actually a square, which sides are of the length \(3\).

We must make a one litre measure, which is shaped as a right circular cylinder without a lid. The problem is to find the best size of the bottom and the height so that we need the least possible amount of material to make the measure.

Let \(r > 0\) be the radius and \(h > 0\) the height of the cylinder. The volume of the cylinder is \(1\) dm\(^3\) and we can write \(\pi r^2 h = 1\) from which we get \[h = \frac{1}{\pi r^2}.\]

The amount of material needed is the surface area \[A_{\text{bottom}} + A_{\text{side}} = \pi r^2 + 2 \pi r h = \pi r^2 + \frac{2 \pi r}{\pi r^2} = \pi r^2 + \frac{2}{r}.\]

Let function \(f: (0, \infty) \to \mathbb{R}\) be defined by \[f(r) = \pi r^2 + \frac{2}{r}.\] We must find the minimum value for function \(f\), which is continuous and differentiable, when \(r>0\). Using the reciprocal rule, we get \[f'(r) = 2\pi r -2 \cdot \frac{1}{r^2} = \frac{2\pi r^3 - 2}{r^2}.\] Now \(f'(r) = 0\), when \[\begin{aligned}2\pi r^3 - 2 &= 0 \\ 2\pi r^3 &= 2 \\ r^3 &= \frac{1}{\pi} \\ r &= \frac{1}{\sqrt[3]{\pi}}.\end{aligned}\]

Let's draw a table:

As the function \(f\) is continuous, we now know that it gets its minimum value at the point \(r= \frac{1}{\sqrt[3]{\pi}} \approx 0.683\). Then \[h = \frac{1}{\pi r^2} = \frac{1}{\pi \left(\frac{1}{\sqrt[3]{\pi}}\right)^2} = \frac{1}{\frac{\pi}{\pi^{2/3}}} = \frac{1}{\sqrt[3]{\pi}} \approx 0.683.\]

This means that it would take least materials to make a measure, which is approximately \(2 \cdot 0.683\) dm \( = 1.366\) dm \( \approx 13.7\) cm in diameter and \(0.683\) dm \( \approx 6.8\) cm high.

Example

Compare the graph of \(\sin x\) (red) with the graphs of the polynomials \[ x-\frac{x^3}{3!}+\frac{x^5}{5!}-\dots + \frac{(-1)^nx^{2n+1}}{(2n+1)!} \] (blue) for \(n=1,2,3,\dots,12\).

If \(f\) is \(n\) times differentiable at \(x_0\), then the Taylor polynomial has the same derivatives at \(x_0\) as the function \(f\), up to the order \(n\) (of the derivative).

The reason (case \(x_0=0\)): Let \[ P_n(x)=c_0+c_1x+c_2x^2+c_3x^3+\dots +c_nx^n, \] so that \begin{align} P_n'(x)&=c_1+2c_2x+3c_3x^2+\dots +nc_nx^{n-1}, \\ P_n''(x)&=2c_2+3\cdot 2 c_3x\dots +n(n-1)c_nx^{n-2} \\ P_n'''(x)&=3\cdot 2 c_3\dots +n(n-1)(n-2)c_nx^{n-3} \\ \dots && \\ P^{(k)}(x)&=k!c_k + x\text{ terms} \\ \dots & \\ P^{(n)}(x)&=n!c_n \\ P^{(n+1)}(x)&=0. \end{align}

From these way we obtain the coefficients one by one: \begin{align} c_0= P_n(0)=f(0) &\Rightarrow c_0=f(0) \\ c_1=P_n'(0)=f'(0) &\Rightarrow c_1=f'(0) \\ 2c_2=P_n''(0)=f''(0) &\Rightarrow c_2=\frac{1}{2}f''(0) \\ \vdots & \\ k!c_k=P_n^{(k)}(0)=f^{(k)}(0) &\Rightarrow c_k=\frac{1}{k!}f^{(k)}(0). \\ \vdots &\\ n!c_n=P_n^{(n)}(0)=f^{(n)}(0) &\Rightarrow c_k=\frac{1}{n!}f^{(n)}(0). \end{align} Starting from index \(k=n+1\) we cannot pose any new conditions, since \(P^{(n+1)}(x)=0\).

Example

Which polynomial \(P_n(x)\) approximates the function \(\sin x\) in the interval \([-\pi,\pi]\) so that the absolute value of the error is less than \(10^{-6}\)?

We use Taylor's Formula for \(f(x)=\sin x\) at \(x_0=0\). Then \(|f^{(n+1)}(c)|\le 1\) independently of \(n\) and the point \(c\). Also, in the interval in question, we have \(|x-x_0|=|x|\le \pi\). The requirement will be satisfied (at least) if \[ |E_n(x)|\le \frac{1}{(n+1)!}\pi^{n+1} < 10^{-6}. \] This inequality must be solved by trying different values of \(n\); it is true for \(n\ge 16\).

The required approximation is achieved with \(P_{16}(x)\), which fo sine is the same as \(P_{15}(x)\).

Check from graphs: \(P_{13}(x)\) is not enough, so the theoretical bound is sharp!

Taylor polynomial and extreme values

If \(f'(x_0)=0\), then also some higher derivatives may be zero: \[ f'(x_0)=f''(x_0)= \dots = f^{(n)}(x_0) =0,\ f^{(n+1)}(x_0) \neq 0. \] Then the behaviour of \(f\) near \(x=x_0\) is determined by the leading term (after the constant term \(f(x_0)\)) \[ \frac{f^{(n+1)}(x_0)}{(n+1)!}(x-x_0)^{n+1}. \] of the Taylor polynomial.

This leads to the following result:

Newton's method

The first Taylor polynomial \(P_1(x)=f(x_0)+f'(x_0)(x-x_0)\) is the same as the linearization of \(f\) at the point \(x_0\). This can be used in some simple approximations and numerical methods.

The recursion formula is based on the geometric idea of finding an approximative zero of \(f\) by using its linearization (i.e. the tangent line).

Example

Find an approximate value of \(\sqrt{2}\) by using Newton's method.

We use Newton's method for the function \(f(x)=x^2-2\) and initial value \(x_0=2\). The recursion formula becomes \[ x_{n+1}= x_n-\frac{x_n^2-2}{2x_n} = \frac{1}{2}\left(x_n+\frac{2}{x_n}\right), \] from which we obtain \(x_1=1{,}5\), \(x_2\approx 1{,}41667\), \(x_3\approx 1{,}4142157\) and so on.

By experimenting with these values, we find that the number of correct decimal places doubles at each step, and \(x_7\) gives already 100 correct decimal places, if intermediate steps are calculated with enough precision.

Taylor series

The Taylor series can be formed as soon as \(f\) has derivatives of all orders at \(x_0\) and they are substituted into this formula. There are two problems related to this: Does the Taylor series converge for all values of \(x\)?

Answer: Not always; for example, the function \[ f(x)=\frac{1}{1-x} \] has a Maclaurin series (= geometric series) converging only for \(-1 < x < 1\), although the function is differentiable for all \(x\neq 1\): \[ f(x)=\frac{1}{1-x} = 1+x+x^2+x^3+x^4+\dots \]